Physics: Principles with Applications (7th Edition)

Published by Pearson
ISBN 10: 0-32162-592-7
ISBN 13: 978-0-32162-592-2

Chapter 15 - The Laws of Thermodynamics - Problems - Page 439: 31

Answer

0.15.

Work Step by Step

We note that 55 mph is 24.582 m/s. Calculate the energy per second (power) needed to overcome the drag forces. $$P_{car}=Fv=(350N)( 24.582 m/s)=8604W$$ Calculate the heat energy per second (power) delivered by the gasoline. This is the energy content of a gallon of gas, divided by the time it lasts. A car that gets 32 mph while traveling at 55 mph will burn one gallon of gasoline in 32/55 hours, or 2094.45 seconds. $$P_{gas}=\frac{(3.2\times10^7J/L)(3.8L/gallon)}{2094.45 s }=58056W$$ Equation 15-4a says that the ratio of these is the efficiency. $$e=\frac{W}{Q_H}=\frac{W/t}{Q_H/t}=\frac{8604W}{58056}=0.148\approx 0.15$$
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